Active current for SOZ

Nelson Pass

The one and only
Paid Member
2001-03-29 12:38 am
You can replace the 8 ohm resistors on the
negative side with a constant current source
and lower the negative voltage rail to 15 volts
or so, but if you put constant current sources
in place of the 8 ohm resistors on the positive
side, you'll have to come up with some way to
stabilize the output, either with feedback or
using load resistance.
 
Vince,

I see you added the bjt biassing transistors to your schematic. I think I will have a crack at building this design in the very near future (I have all the necessary parts).

I too was confused by Nelson's comment about 15V negative rail.

Nelson,

Does this mean the gain in efficiency is due to the ability to use a smaller negative rail voltage? Could you elaborate on this at all.

Also it was said in a previous post (I don't remember by whom) that the third harmonic distortion of the SOZ is due to the use of biassing resistors and not CCS's as in the ZEN, can someone explain or correct this statement please.

Cheers

Dan
 

Nelson Pass

The one and only
Paid Member
2001-03-29 12:38 am
Regardless of power, if you use a current
source on the negative side of SOZ, you
need to allow for about 5 volts drop (Vgs)
of the main gain devices, the probable
input voltage variation, which in balanced
mode is very small but is equal to peak input
signal for single-ended input, and the minimum
voltage required to operate the current source.
This comes to about 10-15 volts.

The main advantage besides lower dissipation is
better performance with a single-ended input, and
better PSRR against the negative rail.
 

vdi_nenna

Member
Paid Member
2000-10-10 7:27 pm
PA, USA
That's good to hear, because I have BOZ as a preamp, until I build the balanced line stage.

Calculating the output power would be more like for the original Zen? 20% of total power dissipated? As opposed to 5% for original SOZ.

Should the CCS MOSFETs be matched also? It this critical?

I'm going to try it very soon also. I need a few more parts.

Thanks!

Vince
 
Trying to understand

Vince:
I got your schematic off your web site. I am interested in seeing how you implement it. I noticed in the Zen article that the drain of Q2 drops to 17 v. I dont understand about dropping the volts on the - rail. Is this already compensated for in your circuit or am I just not getting it .
 

vdi_nenna

Member
Paid Member
2000-10-10 7:27 pm
PA, USA
I don't understand it completely myself. But I have the original artical in front of me now. I think you mean the center drain pin of Q1 is +17v. That's about half the +34v input. Why? I'm not sure, but I bet that's part of the answer in Nelson's last post. Vgs and so on.

We're dealing w/ the CCS on the negative side. (I didn't opt for the + side CCSs, so I wouldn't need to deal w/ the loop.) Truthfully, I dont know the answer. Sorry. I'm running on some blind faith here. :)

Vince
 
Starting to see

Vince:
What Nelson is alluding to (I think?) is a combination of the input signal(as it relates to the voltage swing of its corresponding gain device)+ the minimum voltage required to power your ccs circuit. He is saying to allow 5v for the input device + 5-10v for the ccs. I wonder if your higher voltage on the - rail would overdrive the ccs or not. Doesnt appear to in the Zen amp. Does that make any sense.
 
Vince,
Remember when I mentioned asymmetrical rails back when we were talking about current sources for the SOZ? Same thing Nelson's talking about.

Okay, here's the deal...
You're familiar with the idea of a voltage regulator, right? An ideal voltage regulator (as opposed to a real world one) will deliver any arbitrary amount of current from 0 amps to infinite amps while maintaining an absolutely steady voltage. That's why it's a *voltage* regulator--the voltage remains constant, no matter what, but the current varies.
Now, a current source has another name: current regulator. What's a current regulator do? (You can already see where I'm going, I'll bet.) Locks the current...and lets the voltage vary. Kind of like an upside-down voltage regulator.
Suppose we were to say we wanted a steady 1 amp. We then build a current source with this in mind. How would it behave? If you give a 1A current source a 1 ohm load, it will develop exactly 1V across the load. Simple application of Ohm's Law: I*R=E...1A*1 ohm=1V. But what happens if you give it a 2 ohm load? It will do whatever it has to do to force 1A through the load. In this case, it will develop 2V of output. 1A*2 ohms=2V. An ideal current source could develop 1kV across a 1k resistor, simply because you told it to deliver 1A, no matter what.
So let's consider what happens if you put one into a differential circuit (i.e. the SOZ). When there's no signal, each device will conduct half of the current. If a balanced signal is applied, one device will conduct more at the same time as the other is conducting less, but always such that the instantaneous total is 1A, regardless of whether the ratio is 90/10, 70/30, or 20/80 at any given moment. The current regulator *will* continue to push one amp through the load.
If you put an unbalanced signal in, the device receiving the signal will behave in accordance with the signal--just like you'd expect. But...that current source is a stubborn l'il critter. It *will* have its way. As the device receiving the signal decreases conductance, the rest of the current gets forced through the other side, willy-nilly, thus creating a signal through that device, even though there wasn't any input at the gate (note that the same principles hold for tubes and bipolars, I'm just using MOSFETs because that's how Nelson laid out the SOZ). An important feature to note here is that the signal through the other side of the differential is *out of phase* from the original signal, creating a balanced signal, where none was before. As the first side goes negative, the second side goes positive, and vice versa.
Due to the fact that current sources vary voltage so easily, this leads us to the possibility of asymmetrical rails for the SOZ. If you program a current source for 1A and give it, say, a -30V rail, it will develop any voltage it has to in order to supply 1A to the load and--this is the keen part--it's self adjusting! If you give it a -10V rail, it will adjust. If you give it a -300V rail, it will adjust. All because a current source excels at changing voltage whilst maintaining current.
Since the SOZ circuit doesn't need a whole lot of negative voltage underneath the MOSFET's source pin (we're assuming N-channel devices, here), you can get away with trimming the fat. Leave some wiggle room for the signal, add a volt or two for the current source itself, add bias, and stir, and presto! you've got a recipe for an amp with asymmetrical rails. Set things up to where you've got 10-15V under the tail of the thing, and you can put any voltage you like on top, whether it's 10, 20, or 50V. But the lower rail can stay the same.
There's one caveat: Real world current sources ain't perfect. They suffer from capacitance and other woes that slow them down, making them somewhat less than the ideal solution.

Grey

P.S.: And don't nobody start griping about the difference between current sources and current sinks, and such. I'm using the term current source in the vernacular sense because that's what everybody calls 'em. Okay? Start messin' with me about nomenclature an' Santa Clause won't bring you that order of MOSFETs you want for Christmas!
 
Beggining to understand

Grey:
When you said " leave some wiggle room for the signal ,add a volt or two for the current source itself ,add bias and presto" does the bias come from the 8ohm and one ohm resistors? Im not trying to sound stupid just ignorant but im slowly starting to understand something about electronics and its exciting. I can generally build from someone elses circuit but im blind to the theory.For some reason this is starting to make sense. Thanks for your time and input.
 

vdi_nenna

Member
Paid Member
2000-10-10 7:27 pm
PA, USA
Grey,

Excellent explaination! One last things. ;)

The voltage on the neg. side should stay inside the rating of the MOSFET? Or is best to keep it around -15v.

The reason I'm asking is because it makes it a bit hard when it comes time to make the power supply. It would be nice to use a +20/-20 or +30/-30 supply. (I could use a larger supply then use a dividing network to bring the voltage to -15). The Fets I chose are 50 volt types, BTW.

Second, it sounds like the 2 sides are not completely independent, e.g., the horizontal 1 ohm power resistor. Being the current should be the same, matching the CCS FETs is necessary. Is this correct?

Thanks a bunch!

Vince
 
Vince,
Working in reverse order:
Rather than match the devices and all the attendant malarky, it would be easier to trim the sense resistor with a pot while measuring the current across the 1 ohm resistor above the I-source. That way you could balance all the variables at once (including, in theory, any remaining mismatch between the gain devices) without having to buy a whole box of semiconductors.
50V MOSFETs sound kinda scant. How much wattage were you shooting for? A +-30V rail would be 60V, top to bottom. Yes, a decent amount of that would fall across the top 8 ohm resistors, but shooting from the hip (and not knowing which devices you've got in mind) I'd guess that you're likely to be run awfully close to the boundaries of the SOA chart (Safe Operating Area...it'll be in the specs for the device).
Yes, to the extent that the MOSFET's source will run 4 or 5 volts down into the negative voltage domain, you'll have to include that voltage as part of what the MOSFET will have to withstand.

Grey
 
Vince,
You must like living dangerously.
Have you already bought the MOSFETs? If so, you might--just possibly--consider using them for the current sources and buy some with somewhat higher voltage ratings for the gain devices. If you haven't bought them, you might shoot for something more like 100V (maybe more) rating. Why? Because we live in a cold, cruel world where things don't always *quite* measure up to specs, and you're already going to be pretty close to the edge.
That said, you might be interested in the idea that I've got a baby SOZ up and running. By baby, I mean single-digit wattages, and that's pushing it. It's useless to me, as-is, because it doesn't have enough power.
So why build it?
Well, you see, it's a guinea pig. I've got plans for the little booger. One thing I'm going to do is put current sources under it. Another thing is to mate the BOSOZ to it in order to up the gain (pretty low for a stock SOZ) and provide an on-board phase splitter. (Yes, for those who think in terms of such things, a tube front end would be a trivial substitution. I'll eventually get around to trying some 6SN7s or 6922s...) But that's nothing more than a warm-up for the real stuff. There are at least two other things I'm going to try before all is said and done.
What things?
That'd be telling...an' it's a secret.

Grey

(Hint for one: More power, less device dissipation...)
(Hint for the other: Blending in another of Nelson's ideas...)